Kahou Wong, On Semiconductor
A buck/boost converter can step a voltage up or down. Such a converter is appropriate for battery-powered applications. One application derives a regulated 14.1 V at 1 A from 12 V solar panels with 9 to 18 V variation. In this type of battery application, efficiency is an important factor; hence, this design uses an inexpensive synchronous-rectifier-based MC33166/7 circuit. It is difficult to find a buck-boost controller in the market. It is even more difficult to find an inexpensive one with an integrated high-current switch. One way to build a buck-boost converter is to use a buck regulator with an internal switch, such as the MC33166/7 (Figure 1). The negative-polarity output voltage connects to the IC's ground pin, and a resistor pair divides the 14.1 V output voltage to 5 V for connection to the FB pin of the IC. So, the IC effectively regulates an input of VIN + VOUT = 18 + 14.1 = 32.1 V to an output of 14.1 V.
|Figure 1.||This inexpensive buck-boost controller uses synchronous
rectification for high efficiency.
The MC33166/7 has a 40 V maximum switch rating, and it can accommodate 95% duty cycle, so it's adequate for the application. To implement synchronous rectification for better efficiency, the design uses an additional transformer winding and a MOSFET. The auxiliary winding provides bias voltage to turn on the MOSFET when the switching-node polarity turns negative. Note that the synchronous rectifier is an important factor in the efficiency of this circuit, because the input-to-output ratio is approximately 1-to-1. So, the duty cycle is approximately 50%, which means that the MOSFET conducts for half the switching-frequency period.
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